a question is in details?

Question

a 0.50kg steel ball travelling at 20m/s strikes very large block and rebounds the same speed. the ball and block deform during the impact allowing the ball's center of mass to move only one tenth milimeter.
A) what is the acceleration of ball during the impact?
if you know this problem, please do it.
thank you

Answer 1

Let's start with an extreme simplification: a small cube sliding into a large block and rebounding at the same speed.

Consider the case where the cube acts (deforms) as a simple spring.

http://en.wikipedia.org/wiki/Hooke's_law

The initial energy of the cube is given by (1/2)MV^2. When the spring has been deflected a distance X, the energy in the spring is (1/2)KX^2.

If we assume that the deformation is uniform over the entire cube and that the block itself does not deform, then X is twice the distance that the center of the cube moved.

Since we have M, V, and X, we can compute K.

The spring equation for force is:
F = -KX

so knowing K and X gives us the peak force and F = MA gives us the peak acceleration.

But the acceleration is not constant. Computing it as a function of the displacement is easy - it is just:
A = -KX/M

If we have to compute it as a function of time then we can use the differential equations of motion to do so:

A(t) = -KX(t)/M
X(0) = 0 (no deflection when just touching)

The result is a sinusoid:
http://en.wikipedia.org/wiki/Simple_harmonic_motion

This gives us both the total rebound time (1/2 cycle or (pi)(sqrt(M/K))) and the acceleration as a function of time.

Now, what changes when we have a ball rather than a small cube?

1. The spring behavior is more complicated.

2. If the ball is rolling rather than sliding, then it also has rotational kinetic energy when it hits the block, and this affects (increases) the energy stored in the compression (i.e. deformation) and hence the peak acceleration.

The above assumed that "spring constant" of the block was essentially infinite. In fact, to first order this doesn't matter. The spring that is the block is acting in parallel with the spring that is the cube/ball so its behavior is already accounted for.

I don't know if this really solves the problem you've been assigned, but I hope it helps.

Answer 2

actually the velocity of the ball changes from v = 20 m/s to 0, then to -20 m/s
during v to 0, the kinetic energy 1/2*mv^2 decreases because of the work done by the average force F that is -F*d where d = 1/10 mm = 10^(-4) m
so F*d = 1/2*mv^2, F = mv^2/(2d)
F = ma, so a = v^2/(2d) = 20^2/10^(-4) m/s^2 = 4.0 * 10^5 m/s^2